April 25, 2019, 02:37:55 pm

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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #17655 on: February 11, 2019, 09:16:22 pm »
0
Are you aware that $\frac{b-c}{c-b} = -1$?

For the next one, use the fact that $m = \tan(\theta)$.
Wait is there a formula for that to be -1? Or did you just try random numbers? How did you work that out?
If so
-1 = $\tan(\theta)$
I'm also unsure what this turns to now >.< cause I'm not familiar with rearranging trig functions
WAIT SORRY
I know now cause it was just the wrong way around so i didnt understand at that point
Modification
$\tan\theta=-1$
$\tan^{-1}\left(-1\right)=\theta$
= -45

Had a look at and the answer was 135 degrees, is that supposed to be correct cause 180-45 = 135 but i'm not sure if there is such a formula
« Last Edit: February 11, 2019, 09:20:22 pm by aspiringantelope »
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#### darkz

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##### Re: VCE Methods Question Thread!
« Reply #17656 on: February 11, 2019, 09:27:59 pm »
+1
Wait is there a formula for that to be -1? Or did you just try random numbers? How did you work that out?
If so
-1 = $\tan(\theta)$
I'm also unsure what this turns to now >.< cause I'm not familiar with rearranging trig functions
WAIT SORRY
I know now cause it was just the wrong way around so i didnt understand at that point
Modification
$\tan\theta=-1$
$\tan^{-1}\left(-1\right)=\theta$
= -45

Had a look at and the answer was 135 degrees, is that supposed to be correct cause 180-45 = 135 but i'm not sure if there is such a formula

Well I never really fully understood these questions haha, but then I'd just assume that they they're looking for a positive answer. So then yeh, that'd be 135 (Refer to the attached diagram)

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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #17657 on: February 11, 2019, 09:32:29 pm »
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Well I never really fully understood these questions haha, but then I'd just assume that they they're looking for a positive answer. So then yeh, that'd be 135 (Refer to the attached diagram)
Ok thanks! Just wondering where did the -(b-c)/b-c (I thought it was c-b/b-c as well o.O) (not b-c/b-c) (-) negative sign come from? the one in front of b-c on the numerator.
Thanks
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#### MB_

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##### Re: VCE Methods Question Thread!
« Reply #17658 on: February 11, 2019, 09:37:13 pm »
+1
Ok thanks! Just wondering where did the -(b-c)/b-c (I thought it was c-b/b-c as well o.O) (not b-c/b-c) (-) negative sign come from? the one in front of b-c on the numerator.
Thanks
-(b-c) is equal to c-b it's just writing it as -b+c and taking out -1 to get -(b-c)
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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #17659 on: February 11, 2019, 09:49:42 pm »
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-(b-c) is equal to c-b it's just writing it as -b+c and taking out -1 to get -(b-c)

Where did the -1 come from o.O?
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#### darkz

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##### Re: VCE Methods Question Thread!
« Reply #17660 on: February 11, 2019, 09:52:54 pm »
+1
Where did the -1 come from o.O?

You factor it out, its just like
$2(x+2)=2x+4$

$-(a+b)=-a-b$

$-1\times a + -1\times b = -1(a+b)$
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#### EllingtonFeint

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##### Re: VCE Methods Question Thread!
« Reply #17661 on: February 12, 2019, 02:52:06 am »
0
Hello.
How do you get your Ti-nspire Cx Cas calculator out of Set-to-test mode?!!
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#### C14M8S

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##### Re: VCE Methods Question Thread!
« Reply #17662 on: February 12, 2019, 09:14:21 am »
0
Hey guys, I've got two questions -
1) Is it possible to map transformations from a matrix onto a function using the TI-nspire?
2) How would I work through the questions that I've attached? I understand how matrix transformations work, but I'm struggling to work through mapping inverse functions.
Cheers!
« Last Edit: February 12, 2019, 09:42:07 am by C14M8S »
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#### S_R_K

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##### Re: VCE Methods Question Thread!
« Reply #17663 on: February 12, 2019, 12:02:32 pm »
+1
Hello.
How do you get your Ti-nspire Cx Cas calculator out of Set-to-test mode?!!

You'll need to connect it to a computer with the Ti-nSpire CAS software, and there's an option to exit press-to-test mode. Ask your teacher.

Hey guys, I've got two questions -
1) Is it possible to map transformations from a matrix onto a function using the TI-nspire?

Yes, but i think it's probably more trouble than its worth.

Quote
2) How would I work through the questions that I've attached? I understand how matrix transformations work, but I'm struggling to work through mapping inverse functions.
Cheers!

For the first one (the second one is pretty much the same), find x' and y' by multiplying that 2x2 matrix by [x y] (as a column matrix, of course). That will give x' = 3y and y' = –2x. Then solve for y in terms of x', x in terms of y', and substitute into the original equation to get (x'/3) = 2(-y'/2) + 3. Rearrange to y = mx + c form.

#### C14M8S

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##### Re: VCE Methods Question Thread!
« Reply #17664 on: February 12, 2019, 04:22:56 pm »
0
Hey again,

How would I use the chain rule to differentiate the following? I couldn't really see how it ties into the problem, when I took g'(x) I got a result of 1/tan(x) and then simplified g'(pi/4) to 1/1 = 1, but the examiner's report says otherwise. Could someone please point out why I'm wrong?

« Last Edit: February 12, 2019, 05:18:57 pm by C14M8S »
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#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #17665 on: February 12, 2019, 05:15:09 pm »
0
Hey again,

How would I use the chain rule to differentiate the following? I couldn't really see how it ties into the problem, when I took g'(x) I got a result of 1/tan(x) and then simplified g'(pi/4) to 1/1 = 1, but the examiner's report says otherwise. Could someone please point out why I'm wrong?

have you attached the right question? It seems to not be relevant.

#### C14M8S

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##### Re: VCE Methods Question Thread!
« Reply #17666 on: February 12, 2019, 05:20:19 pm »
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have you attached the right question? It seems to not be relevant.
Thanks for pointing that out - I've attached the correct question now >.<
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#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #17667 on: February 12, 2019, 05:37:24 pm »
+2
$g(x) = log((tan(x)))$$g'(x) = \frac{sec^2(x)}{tan(x)}$$g'(\frac{\pi}{4}) = \frac{sec^2(\frac{\pi}{4})}{tan(\frac{\pi}{4})}$$g'(\frac{\pi}{4}) = \frac{(\sqrt{2})^2}{1} = 2$

Just make sure that when you are differentiating logs the structures is if g(x) = log(f(x)) then g'(x) = f'(x)/f(x)

#### EllingtonFeint

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##### Re: VCE Methods Question Thread!
« Reply #17668 on: February 13, 2019, 07:54:53 pm »
0
I’ve just started learning inverse functions (I’m not behind, am I?!) and I don’t really understand how to find the domain and range of an inverse so could somebody please explain it to me?
So for example...

For the following function find the inverse and state it’s domain and range
f:[-2,0] —> R, f(x) = 2x - 4
So, I found that the inverse is...
f^-1 (x) = 1/2 (x+4)

I’m just not sure how to find the domain and range.
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#### EllingtonFeint

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##### Re: VCE Methods Question Thread!
« Reply #17669 on: February 13, 2019, 07:55:40 pm »
0
I’ve just started learning inverse functions (I’m not behind, am I?!) and I don’t really understand how to find the domain and range of an inverse so could somebody please explain it to me?
So for example...

For the following function find the inverse and state its domain and range
f:[-2,0] —> R, f(x) = 2x - 4
So, I found that the inverse is...
f^-1 (x) = 1/2 (x+4)

I’m just not sure how to find the domain and range.
2018: Biology [49] | Indonesian SL: [40]

🌱Offering biology tutoring in Mornington Peninsula area or online 🌱